Eliminating the parameter from a parametric equation

There are three ways to eliminate the parameter from a parametric equation

Given a parametric curve where our function is defined by two equations, one for $x$ and one for $y$, and both of them in terms of a parameter $t$,

$x=f(t)$

$y=g(t)$

we can eliminate the parameter value in a few different ways.

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We can

1. solve each equation for the parameter $t$, then set the equations equal to one another, or

2. solve one equation for the parameter $t$, then plug that value into the second equation, or

3. solve each equation for part of an identity, then plug both values into the identity.

How to eliminate the parameter

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Eliminating the parameter using the second method

Example

Eliminate the parameter.

$x=2t^2+6$

$y=5t$

We’ll solve $y=5t$ for $t$, since this will be easier than solving $x=2t^2+6$ for $t$.

$y=5t$

$t=\frac{y}{5}$

Plugging this into the equation for $x$, we get

$x=2\left(\frac{y}{5}\right)^2+6$

$x=\frac{2y^2}{25}+6$

Removing the fraction, we get

$25x=2y^2+150$

$25x-2y^2=150$

Given a parametric curve where our function is defined by two equations, one for x and one for y, and both of them in terms of a parameter t, we can eliminate the parameter in a few different ways.

Example

Eliminate the parameter.

$x=e^t$

$y=e^{4t}$

We know that $y=e^{ab}$ is the same as $y=(e^a)^b$. If we use this property, we can take $y=e^{4t}$ and rewrite it as $y=(e^t)^4$. Since $x=e^t$, we can substitute $x$ into $y=(e^t)^4$ for $e^t$.

$y=x^4$

Remember, because we have $e$ in the original parametric equations, and $e$ requires that $t>0$, we have to transfer this condition to our final answer, and say

$y=x^4$, where $x>0$

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Let’s try another example using the third method.

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Example

Eliminate the parameter.

$x=2\cos{\theta}$

$y=3\sin{\theta}$

$0\le\theta\le2\pi$

Rearranging $x=2\cos{\theta}$ and $y=3\sin{\theta}$ to isolate the trigonometric functions, we get

$x=2\cos{\theta}$

$\cos{\theta}=\frac{x}{2}$

and

$y=3\sin{\theta}$

$\sin{\theta}=\frac{y}{3}$

Since we know that $\sin^2{\theta}+\cos^2{\theta}=1$, we can substitute the values we just found for $\cos{\theta}$ and $\sin{\theta}$.

$\left(\frac{y}{3}\right)^2+\left(\frac{x}{2}\right)^2=1$

$\frac{y^2}{9}+\frac{x^2}{4}=1$

$y^2+\frac{9x^2}{4}=9$

$4y^2+9x^2=36$

$9x^2+4y^2=36$

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